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Current demand of battery charger?

P

Patrik

Jan 1, 1970
0
Hello

I'm designing a battery charger for NiMh cells, four of them to be
exact. One way of designing this is to use _one_ current source that
feeds all four cells that are connected in series. Another way of
design is using four separate current sources, on for each cell. This
second way seems most interresting because of the capability to
measure the temperature of each cell and stop loading just the actual
cell that is fully charged.
I decided to load invidually, and began to think about how much
current this charger will require.
Say I use only one current source, set to 1.5 ampere. The charger then
needs just 1.5A, and the GP2000 cells will be fully charged in
(estimated) 1 hour and a half.
Say instead I use four parallel current sources, set to 1.5A. The
charger then needs 6A, and the GP2000 cells will be fully charged in
(again estimated) 1.5 hour.

Or?

This doesn't seem correct, can somebody please tell me what is
correct?

Regards Patrik
 
N

Nathan Schemm

Jan 1, 1970
0
[email protected] (Patrik) wrote in message
Say I use only one current source, set to 1.5 ampere. The charger then
needs just 1.5A, and the GP2000 cells will be fully charged in
(estimated) 1 hour and a half.
Say instead I use four parallel current sources, set to 1.5A. The
charger then needs 6A, and the GP2000 cells will be fully charged in
(again estimated) 1.5 hour.

First of all, I am new in electronics as well but I think I can help
you. You are correct in the way you are looking at the current draw.
One thing that might help you is if you look at the total input watts
in each case. In each case the Volts times the Amps is the same. So
the same power is applied in both cases. Using the lower voltage,
higher amperage meathod may make it harder to build an efficient power
supply, but if the efficiency of the power supply in both cases is
constant, the same power will be drawn from the source.

Hope that helps
Nathan
 
G

Gary Lecomte

Jan 1, 1970
0
Hello

I'm designing a battery charger for NiMh cells, four of them to be
exact. One way of designing this is to use _one_ current source that
feeds all four cells that are connected in series. Another way of
design is using four separate current sources, on for each cell. This
second way seems most interresting because of the capability to
measure the temperature of each cell and stop loading just the actual
cell that is fully charged.
I decided to load invidually, and began to think about how much
current this charger will require.
Say I use only one current source, set to 1.5 ampere. The charger then
needs just 1.5A, and the GP2000 cells will be fully charged in
(estimated) 1 hour and a half.
Say instead I use four parallel current sources, set to 1.5A. The
charger then needs 6A, and the GP2000 cells will be fully charged in
(again estimated) 1.5 hour.

Or?

This doesn't seem correct, can somebody please tell me what is
correct?

Regards Patrik

Yes, that is correct.
And charging at that rate, it is really important to monitor
temperature or better yet the point at which the cell voltage starts
to decline after increasing to peak.

Take care...Gary
 
R

Rich Grise

Jan 1, 1970
0
[email protected] (Patrik) wrote in message > Hello
....
I decided to load invidually, and began to think about how much
current this charger will require.
Say I use only one current source, set to 1.5 ampere. The charger then
needs just 1.5A, and the GP2000 cells will be fully charged in
(estimated) 1 hour and a half.
Say instead I use four parallel current sources, set to 1.5A. The
charger then needs 6A, and the GP2000 cells will be fully charged in
(again estimated) 1.5 hour.

Or?

This doesn't seem correct, can somebody please tell me what is
correct?

Regards Patrik

What doesn't seem correct about it? 1.5 + 1.5 + 1.5 + 1.5 = 6, after all. : )

Cheers!
Rich
 
R

Rich Grise

Jan 1, 1970
0
[email protected] (Patrik) wrote in message
....
First of all, thankyou for your answers, that about total power and
watts made it somehow clearer. But what made me think about it from
the beginning was really this thing with electric charge, that is how
many electrons gets in the cells. The case when I use four current
sources must put more electrons in the cells, which should mean the
cell is charged more quickly?

Imagine a sort of trough, or channel, full of ping-pong balls. (single-
file). At the end, they fall off and into a bucket. Say it takes 100
ping-pong balls to fill one bucket. Simple enough, right? So, now,
add 3 buckets, and put a fork in the channel, so you're filling up
four buckets. How many ping-pong balls do you need?

Good Luck!
Rich
 
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